Question

A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zero charge. How long after the switch is closed will it take the capacitor to reach 2/3 its maximum charge?

Answer 1

t = 1.098*RC

Step-by-step explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

`Q=Q_(max)[1-e^{-(t)/(RC)}]` (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

`Q=(2)/(3)Q_(max)=Q_(max)[1-e^{-(t)/(RC)}]\\\\(2)/(3)=1-e^{-(t)/(RC)}\\\\e^{-(t)/(RC)}=(1)/(3)\\\\-(t)/(RC)=ln((1)/(3))\\\\t=-RCln((1)/(3))=1.098RC`

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC